operator-precedence

I've just read that order of evaluation and precedence of operators are different but related concepts in C++. But I'm still unclear how those are different but related?. int x = c + a * b; // 31 int y = (c + a) * b; // 36 What does the above statements has to with order of evaluation. e.g. when I say (c + a) am I changing the order of evaluation of expression by changing its precedence? The important part about order of evaluation is whether any of the components have side effects. Suppose you have this: int i = c() + a() * b(); Where a and b have side effects: int global = 1; int a() {

Almost all C/C++ operator precedence tables I have consulted list the ternary conditional operator as having higher precedence than the assignment operators. There are a few tables, however, such as the one on wikipedia , and the one at operator-precedence.com , that place them on the same precedence level. Which is it, higher or same? In the C++ grammar, assignment-expression: conditional-expression logical-or-expression assignment-operator initializer-clause throw-expression conditional-expression: logical-or-expression logical-or-expression ? expression : assignment-expression initializer

I came across this problem in this website , and tried it in Eclipse but couldn't understand how exactly they are evaluated. int x = 3, y = 7, z = 4; x += x++ * x++ * x++; // gives x = 63 System.out.println(x); y = y * y++; System.out.println(y); // gives y = 49 z = z++ + z; System.out.println(z); // gives z = 9 According to a comment in the website, x += x++ * x++ * x++ resolves to x = x+((x+2)*(x+1)*x) which turns out to be true. I think I am missing something about this operator precedence. Java evaluates expressions left to right & according to their precedence. int x = 3, y = 7, z = 4; x

Why does the first line return TRUE, and the third line returns 1? I would expect both lines to return 1. What is the exact meaning of those extra two parentheses in the third line? !is.na(5) + !is.na(NA) # TRUE (!is.na(5)) + (!is.na(NA)) # 1 edit: should check these multiple times. The original problem was with !is.na() , thought it replicated for is.na() . But it didn't :) ! has a weird, counter-intuitive precedence in R. Your first code is equivalent to !(is.na(5) + !is.na(NA)) That is, ! has lower precedence than + . 来源： https://stackoverflow.com/questions/17651687/behavior-of-summing-is

int i=2; i = ++i + ++i + ++i; Which is more correct? Java's result of 12 or C = 13. Or if not a matter of correctness, please elaborate. There is nothing like more correct. It is actually undefined and its called Sequence Point Error. http://en.wikipedia.org/wiki/Sequence_point Matthew Flaschen Java guarantees ( §15.7.1 ) that it will be evaluated left-to-right, giving 12. Specifically, ++ has higher precedence that + . So it first binds those, then it associates the addition operations left to right i = (((++i) + (++i)) + (++i)); §15.7.1 says the left operand is evaluated first, and §15.7.2

So, the C++ standard requires that class members be initialized in the order in which they are declared in the class, rather than the order that they're mentioned in any constructor's initializer list. However, this doesn't imply anything about the order in which the arguments to those initializations are evaluated. I'm working with a system that frequently passes references to serialization objects around, and wondering if I can ensure that bits are read from it in the right order, independent of the order in which those bits get written into the object's fields. struct Foo { int a; double b;